# Help: Vector field and radius vector

1. Oct 1, 2006

### Mathman23

Hi Guys,

Given the vector field $$X(x,y) = ( a + \frac{b(y^2-x^2)}{(x^2+y^2)^2}, \frac{-2bxy}{(x^2+y^2)^2}}})$$

Show that for a point (x,y) on the circle with radius r = \sqrt(b/a) (i.e. x^2 + y^2 = b/a), the vector X(x,y) is tangent to a circle at the point.

My strategy is that to first define a vector function r(t) and show that this function is to a point on the circle?

Cheers and Best Whishes MM23

p.s. the potential function for X(x,y) is F(x,y) = ax + (bx/x^2+y^2).

p.p.s. Do I first find the directional dereative for F(x,y) ?

And then check to see if this vector is a tangent for the circle?

my solution:

Let

$$r (\theta )=\sqrt{\tfrac{b}{a}}(\cos \theta , \, \sin \theta ), \qquad 0\le \theta <2\pi$$

be any point on the circle. Plugging into the vector field one get at such a point the vector

$$a(1+\sin ^2 \theta -\cos ^2 \theta ,\, -2\cos \theta \sin \theta )=2a (\sin ^2 \theta ,\, -\cos \theta \sin \theta ).$$

One realizes upon a scalar multiplication by the radius vector that this vector is orthogonal to it. Thus...

If this is correct to I then conclude that $$r (\theta ) \cdot X = 0$$ ??

Last edited: Oct 1, 2006
2. Oct 1, 2006

### HallsofIvy

Staff Emeritus
Yes, that works nicely.